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What are the current best asymptotic bounds on $\pi^{-1}(x)$, where $\pi(x)$ denotes the prime counting function (number of primes at most $x$)?

In other words, I am curious about the state of the art for estimating the $n^{th}$ prime. From the prime number theorem, it seems clear that $\pi^{-1}(x)=\Theta(x \log x)$. Can someone point me in the direction of literature that answers the question? Please excuse my inability to find such literature myself...

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en.wikipedia.org/wiki/… $\;$ –  Ricky Demer May 24 '13 at 5:24
Thanks, should I go ahead and delete this question? –  vlv May 24 '13 at 5:59
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3 Answers

up vote 7 down vote accepted

Here is how we invert the prime counting function $\pi(x)$ to estimate the n-th prime $p_n$. Let $g(u),u \ge 2$ be positive, continuous increasing function and let $f(x)$ be defined by

$$ x = \int_{2}^{f(x)} \frac{du}{g(u)} $$


$$ f(x) = \int_{2}^{x} g(f(u))du + f(2) $$

Take $g(u) = \ln u$ so our first estimate of $f(x) is $that $f(x) \sim x\ln x$. Now use this estimate in the above integral to obtain the second estimate for $f(x)$. You can repeat this iterative process $k$ times to obtain the asymptotic expansion of $f(x)$ with an error $O(x/(\ln x)^{-k})$.

Now from the Prime number theorem, we have

$$ \pi(x) \sim \int_{2}^{x} \frac{du}{\ln(u)} $$

Taking $x = p_n$, the n-th prime, the above method says that the asymptotic expansion of $p_n$ is $f(n)$. Thus we obtain

$$ p_n \sim n\ln n + n\ln\ln n - n + \ldots $$

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Assuming the Riemann hypothesis $$|p_n -\text{li}^{-1}(n) |\le \pi^{-1} \sqrt{n} (\log n)^{5/2}, \qquad n \ge 11$$ This follows from known bounds for $\pi(x)$ due to Schoenfeld.

You may see the details in the paper http://front.math.ucdavis.edu/1203.5413

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I like this answer the best, but I'm keeping Nilotpal's answer as the "accepted" one because his is more informative. –  vlv Feb 8 at 6:08
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[I think this is morally the same as Nilotpal Sinha's answer, but it only uses asymptotic estimates and no integrals, so I thought I'd post it here in case someone found it useful or interesting.]

Start with the version of the Prime Number Theorem which says that $\pi(x) \sim x / \ln(x)$. Let $p_n$ denote the $n$-th prime, so in particular $\pi(p_n) = n$, and substitute in $p_n$ in the asymptotic estimate for $\pi(x)$ to get $n = \pi(p_n) \sim p_n / \ln(p_n)$, or equivalently

$p_n \sim n \ln p_n$

(If you're being nitpicky this works because the sequence $p_n$ is monotonic and unbounded.) Now take logs of both sides (again, if you're being nitpicky, this works because both $n$ and $p_n / \ln(p_n)$ are unbounded) and multiply by $n$ to get:

$n \ln p_n \sim n \ln(n) + n\ln\ln(p_n)$.

At this point, if you want just the leading term, you can show easily that $\lim_{n \to \infty} n \ln\ln(p_n) / n\ln(n) = 0$, so that in fact $p_n \sim n \ln p_n \sim n \ln n$, or you can probably keep substituting in to get finer and finer error terms.

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